AP EAMCET Engineering 2013 Solved Paper

Given system of equation is
3x+2y+z=6
3x+4y+3z=14
6x+10y+8z=a
Here, A=[

3	2	1
3	4	3
6	10	8

], B=[

6

14

a

]
C11=(32−30)=2, C12=−(24−18)=−6,
C13=(30−24)=6
C21=−(16−10)=−6, C22=(24−6)=18,
C23=−(30−12)=−18
C31=(6−4)=2, C32=−(9−3)=−6,
C33=(12−6)=6
adj‌A=[

C11	C12	C13
C21	C22	C23
C31	C32	C33

]′ =[

2	−6	6
−6	18	−18
2	−6	6

]′
=[

2	−6	2
−6	18	−6
6	−18	6

]
So, (adj‌A)B=[

2	−6	2
−6	18	−6
6	−18	6

] [

6

14

a

]
=[

12−84+2a

−36+252−6a

36−252+6a

] =[

−72+2a

216−6a

−216+6a

]
and |A|=|

3	2	1
3	4	3
6	10	8

|
=3(32−30)−2(24−18) +1(30−24)
=3(2)−2(6)+6 =6−12+6=0
We know that, if |A|=0 and (adj‌A)‌cdot‌B=0, then the system of equations is consistent and has an infinite number of solutions.
(adj‌A).B=0
⇒ [

−72+2a

216+6a

−216+6a

]=[

0

0

0

]
On comparing, we get
2a−72=0
∴ a=36